Tìm x biết \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
Tìm x biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
Tìm x biết: \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+2\right):2}=\)\(1\frac{2009}{2011}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\left(x+2\right)}{2}}=1\frac{2009}{2011}\)
\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x\left(x+2\right)}=1\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+2\right)}=1\frac{2009}{2011}-1\)
\(\Leftrightarrow\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\right]=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+2}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{2009}{2011}\div2=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Leftrightarrow x=2011-2=2009\)
tìm x biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
Giả sử x là số nguyên dương
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Rightarrow x=2010\)
tìm x, y thỏa mãn
a, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x.\left(x+2\right):2}=1\frac{2009}{2011}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
có \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
tách vế trái đặt là A
ta lại có\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x.\left(x+1\right):2}\right)\)
\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}\)
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)
\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)
\(A=\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}\)
\(A=1+\frac{1}{\left(x+1\right):2}\)
ta thế vào vế trái vào vế phải
ta có\(1+\frac{1}{\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{\left(x+1\right):2}=\frac{2009}{2011}-1\)
\(\frac{1}{\left(x+1\right):2}=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)
\(-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)
thấy hai tử bằng nhau
\(\Rightarrow-\left(x+1\right)=2011\)
\(\Rightarrow\left(x+1\right)=-2011\)
\(\Rightarrow x=-2011-1=-2012\)
Tìm các số nguyên x thỏa mãn:
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+2\right):2}=1\frac{2009}{2011}\)
1+1/3+1/6+1/10+...+1/x.(x+1):2=1+2009/2011
=>2/6+2/12+2/20+...+2/x.(x+1)=2009/2011
=>2.(1/2.3+1/3.4+1/4.5+...+1/x.(x+1))=2009/2011
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=2009/2011:2
=>1/2-1/x+1=2009/4022
=>1/x+1=1/2-2009/4022
=>1/x+1=1/2001
=.x+1=2001
=>x=2001-1
=>x=2000
vậy x=2000
Tìm số tự nhiên x biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
Giúp mk nha làm ơn
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Leftrightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(1-\frac{2}{x+1}=\frac{2009}{2011}\)
\(\frac{2}{x+1}=1-\frac{2009}{2011}\)
\(\frac{2}{x+1}=\frac{2}{2011}\)
\(x+1=2011\)
\(x=2011-1\)
\(\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Rightarrow2\cdot\frac{x-1}{2x+2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{2x-2}{2x+2}=\frac{2009}{2011}\)
Bạn làm nốt.Nhân chéo là ra
\(\left(x-1\right)f\left(x\right)=\left(x+4\right)\cdot f\left(x+8\right)\)
Với \(x=1\) ta có:
\(\left(1-1\right)\cdot f\left(1\right)=\left(1+4\right)\cdot f\left(9\right)\)
\(\Rightarrow5\cdot f\left(9\right)=0\)
\(\Rightarrow f\left(9\right)=0\)
Vậy \(x=9\)
Thay \(x=-4\) vào ta được:
\(\left(-4-1\right)\cdot f\left(-4\right)=0\cdot f\left(4\right)\)
\(\Rightarrow f\left(-4\right)=0\)
Vậy \(x=-4\)
\(\Rightarrow f\left(x\right)\) có ít nhất 2 nghiệm là 9;-4
a) tìm x biết \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
b) cho biết (x-1)f(x)=(x+4)f(x+8) với mọi x
chứng minh f(x) có ít nhất 2 nghiệm